When we see some activities like a running boy, moving bus etc, only one question put in our mind that how these things are moving. For a moving body, speed is the quantity by which we know how fast the body is moving. The concept of motion is described by using some terms like velocity, speed, distance; acceleration etc. motion is described as changed in the position of moving object at an instant of time or period time.

This is one of the basic parameter for motion. It is quite different with speed as it is not only covered length or distance. It is a **scalar quantity** with only magnitude not direction. It was first discovered by *Galileo*. He gave the formula and definition of speed that is distance divided time. It can be calculated in terms of constant, angular, Instantaneous, and average speed. Let’s discuss about speed, its calculation, and its different types.

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Defined as the distance covered by a moving object in unit time taken. It is a scalar quantity and hence it can only be represented by magnitude not by direction.In other words, “**The speed of an object can also be considered as the magnitude of the velocity of the object**”. The speed of a body is the rate of change of distance with time. Numerically it is the distance travelled by the body in 1 sec. It is generally represented by the letter u and v.

Formula :

**Speed = **$\frac{Distance\ covered}{Time\ taken}$** **

**or**

**X =**** **$\frac{d}{t}$ where,

X = speed of the moving object;

d = distance traveled by the moving object;

t = time taken to travel distance ‘d’.

It can be measured in meter per second or m/s or in kmph (kilometer per hour) or fps (feet per second). But the**S. I. unit of speed is m/s**.

It can be calculated by finding how much distance traveled by the body and in how much time with respect to the point of observation. Once we find these two things (distance traveled and time taken to travel the distance), we can divide the distance traveled by the time taken to obtain the speed of the object using formula,

**Speed** = $\frac{d}{t}$.Where

d = distance traveled

t = time taken.

**The Average Speed is defined as** **"the ratio of total distance traveled, irrespective of the direction with that of the time of the observation."**

Let us Consider that the distance traveled by a moving car is denoted by d and it covers this distance in t time, then the average speed**S**_{avg} is given by:

**S**_{avg} = $\frac{d}{t}$

Where

**d** = distance traveled

**t** = time taken.

Although, from the equation it seems that the average speed is same as that of the speed, but there is a significant difference between these two types of speed.

For better understanding of the average speed let’s solve few examples:

### Solved Example

**Question: **A train is traveling from city A to city B with the speed of 70 mph and while traveling back from city B to city A it travels with the speed of 85 mph. Find the average speed of the train for the whole trip?

** Solution: **

As we know that the train is traveling between city A and B, so, the distance traveled by it is two times that of the distance between city A and B.

Now, consider that D = distance between city A and B ,

t_{1} = Time taken by train to travel from city A to B

= $\frac{D}{70}$,

t_{2} = Time taken by train to travel from city B to A

= $\frac{D}{85}$,

Similarly, S_{1}= Speed of the train from city A to B = 70 mph

S_{2} = Speed of the train from city B to A = 85 mph

So, the average speed of the round trip is

S_{avg} = $\frac{(Total\ Distance\ Traveled)}{Time\ Taken}$

$S_{avg}$ = $\frac{2D}{(\frac{D}{70} + \frac{D}{85})}$

= $\frac{(2D \times 70 \times 85)}{155D}$

= $\frac{(2 \times 70 \times 85)}{155}$.

S_{avg} = 76.77mph.

So, the average speed of the train for the round trip is 76.77 mph.

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When a body moves equal amount of distance in equal time interval, the object is said to be moving with the **Constant speed** or **Uniform speed**. An example of the constant speed is the movement of the hour hand and minute hand of the clock.
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**The speed attained by the body at that instant of time. **A speedometer can read the instantaneous velocity of an object. The manual calculation of the instantaneous speed is difficult. We can calculate the instantaneous speed by using distance-time graph.

→ Read More When a object moves along the circular track, it has linear speed as well as the angular speed. The linear is the measure of the distance traveled by the object per unit time. It is the measure of the angle traveled by the object per unit time.

In other words,

**The measure of how fast the angle of the object is changing on the circular track.**
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**Defined as the distance traveled by the object per unit time. **It is denoted by,

V = $\frac{S}{t}$

**$\omega$** by the equation,

V =**$\omega$** r.

It is related to the angular speed $\omega$ by the following relation

S = r**$\theta$**;

Dividing by t, we get,

$\frac{S}{t}$ = r $\frac{\omega}{t}$.V = r $\omega$.

where,**S** = distance traveled,

**$\omega$** = angular speed,

and**r** = radius of the curvature along which object is traveling.
→ Read More
The Speed Problems or commonly known as uniform rate problems involve object moving with either constant speed or with an average. Here you will find “**how far**” or “**how fast**” or “**how long**” the body moves.

### Solved Examples

**Question 1: **If a train is moving at the speed of 40 mph, find the distance it travels in 2.5 s?

** Solution: **

From the question we have,

Speed (X) = 40 m/s

Time (T) = 2.5s

The distance (D) traveled by the train can be given by

Distance = Speed $\times$ Time or

D = X $\times$ T

Substituting the respective values in the above equation, we have,

D = 40 $\times$ 2.5

= 100 miles

So, the distance traveled by the train in 2.5 hours is 100 miles.

**Question 2: **A man drives a car at an average speed of 30 mph to a nearby metro station. He boarded the metro to reach his office which travels at an average speed of 60 mph. The entire distance was 150 miles and the entire trip took three hours. Find the distance from the metro station to his office?

** Solution: **

Let the

From this we can derive three equations and solve to find the various variable; we know that,

Distance = Speed × Time. x = s × t. x = 30t ..............(1)

150 – x = 60 × (3 – t) ............(2) 150 = 3s

So, putting the value of x from (1) in (2) we have (2) as

150 - 30t = 180 - 60t

60t - 30t = 180 - 150

30t = 30

t = 1 hr.

So, the man drive his car for 1 hour which means that he travels for 30 miles through his car in 1 hr. (solving equation 1).

As we know that the total distance traveled by him is 150 miles, So the distance from the metro station to his office is 120 miles (150 - 30).

The Speed of a body is constant,that means speed does not change with time, hence there is no acceleration.

So, speed-time graph for a body moving with constant speed is a straight line parallel to the time axis, i.e., if the speed-time graph of a body is a straight line and parallel to the time axis, the speed of the body is constant.

We know,

Speed = $\frac{Distance\ Traveled}{Time\ Taken}$So, Distance Traveled = Speed $\times$ Time Taken ------(1)

Now, to find out the distance traveled by the body at point C using the graph,

Distance traveled = OA $\times$ OC = Area of rectangle**OABC**.Thus, the area enclosed by the speed-time curve and the time axis gives us the distance traveled by the body.

**Graph When Speed Changes at an Uniform Rate:**

When a body moves with uniform acceleration, its speed changes by equal amounts in equal intervals of time. The graph for an uniformly changing speed will be straight line.

Distance traveled = Area of triangle OPQ

= $\frac{1}{2}$ $\times$ area of rectangle ORPQ

= $\frac{1}{2}$ $\times$ OR $\times$ OQ

When the initial speed of the body is not zero:

The graph given below, shows the graph of a body having an initial speed equal to OB.

The body accelerates from B to C.

Distance traveled = area of OBCA

= $\frac{(Sum\ of\ two\ parallel\ side)(Height)}{2}$

= $\frac{(OB + AC)(OA)}{2}$.

**When speed changes at a non-uniform Rate:**

When the speed of a body changes in an irregular manner, the graph of the body is a curved line. The distance traveled by the body is given by the area between the speed-time curve and the time axis.

From the graph, the distance traveled from point O to point P is equal to the area enclosed by OPR. Distance traveled = Area of OPR.

Formula :

X

d

t

It can be measured in meter per second or m/s or in kmph (kilometer per hour) or fps (feet per second). But the

It can be calculated by finding how much distance traveled by the body and in how much time with respect to the point of observation. Once we find these two things (distance traveled and time taken to travel the distance), we can divide the distance traveled by the time taken to obtain the speed of the object using formula,

d

t

Let us Consider that the distance traveled by a moving car is denoted by d and it covers this distance in t time, then the average speed

Where

Although, from the equation it seems that the average speed is same as that of the speed, but there is a significant difference between these two types of speed.

For better understanding of the average speed let’s solve few examples:

As we know that the train is traveling between city A and B, so, the distance traveled by it is two times that of the distance between city A and B.

Now, consider that D = distance between city A and B ,

t

= $\frac{D}{70}$,

t

= $\frac{D}{85}$,

Similarly, S

S

So, the average speed of the round trip is

S

$S_{avg}$ = $\frac{2D}{(\frac{D}{70} + \frac{D}{85})}$

= $\frac{(2D \times 70 \times 85)}{155D}$

= $\frac{(2 \times 70 \times 85)}{155}$.

S

So, the average speed of the train for the round trip is 76.77 mph.

→ Read More When a object moves along the circular track, it has linear speed as well as the angular speed. The linear is the measure of the distance traveled by the object per unit time. It is the measure of the angle traveled by the object per unit time.

In other words,

V = $\frac{S}{t}$

where, S = distance traveled

t = time taken.

It is related to angular Velocity V =

It is related to the angular speed $\omega$ by the following relation

S = r

Dividing by t, we get,

$\frac{S}{t}$ = r $\frac{\omega}{t}$.V = r $\omega$.

where,

and

From the question we have,

Speed (X) = 40 m/s

Time (T) = 2.5s

The distance (D) traveled by the train can be given by

Distance = Speed $\times$ Time or

D = X $\times$ T

Substituting the respective values in the above equation, we have,

D = 40 $\times$ 2.5

= 100 miles

So, the distance traveled by the train in 2.5 hours is 100 miles.

Let the

Distance | Speed | Time | |

Driving | x | 30 | t |

Traveling by Metro | 150 - x | 60 | 3-t |

Total | 150 | s | 3 |

From this we can derive three equations and solve to find the various variable; we know that,

Distance = Speed × Time. x = s × t. x = 30t ..............(1)

150 – x = 60 × (3 – t) ............(2) 150 = 3s

So, putting the value of x from (1) in (2) we have (2) as

150 - 30t = 180 - 60t

60t - 30t = 180 - 150

30t = 30

t = 1 hr.

So, the man drive his car for 1 hour which means that he travels for 30 miles through his car in 1 hr. (solving equation 1).

As we know that the total distance traveled by him is 150 miles, So the distance from the metro station to his office is 120 miles (150 - 30).

So, speed-time graph for a body moving with constant speed is a straight line parallel to the time axis, i.e., if the speed-time graph of a body is a straight line and parallel to the time axis, the speed of the body is constant.

We know,

Speed = $\frac{Distance\ Traveled}{Time\ Taken}$So, Distance Traveled = Speed $\times$ Time Taken ------(1)

Now, to find out the distance traveled by the body at point C using the graph,

Distance traveled = OA $\times$ OC = Area of rectangle

Distance traveled = Area of triangle OPQ

= $\frac{1}{2}$ $\times$ area of rectangle ORPQ

= $\frac{1}{2}$ $\times$ OR $\times$ OQ

When the initial speed of the body is not zero:

The graph given below, shows the graph of a body having an initial speed equal to OB.

The body accelerates from B to C.

Distance traveled = area of OBCA

= $\frac{(Sum\ of\ two\ parallel\ side)(Height)}{2}$

= $\frac{(OB + AC)(OA)}{2}$.

When the speed of a body changes in an irregular manner, the graph of the body is a curved line. The distance traveled by the body is given by the area between the speed-time curve and the time axis.

From the graph, the distance traveled from point O to point P is equal to the area enclosed by OPR. Distance traveled = Area of OPR.

More topics in Speed | |

Constant Speed | Average Speed |

Instantaneous Speed | Rotational Speed |

Linear Speed | Angular Speed |

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Physics Help | Physics Tutor |